Find a solution for $f\left(\frac{1}{x}\right)+f(x+1)=x$
Not an answer but maybe something to consider for your second functional equation,
Let $\phi$ denote the golden ratio so that we have $\frac{1}{\phi}+1=\phi$
Then by the second functional equation if we set $x=\frac{1}{\phi}$ we have:
$$f(\phi)-f(\frac{1}{\phi}+1)=\frac{1}{\phi}$$ $$f(\phi)-f(\phi)=\frac{1}{\phi}$$ $$0=\frac{1}{\phi}$$
Which obviously isn't true so $f(x)$ isn't properly defined at $x=\phi$
In addition either $f(x)$ isn't analytic at $x=0$ or we must have that:
$$f(x)\sim -x$$
Because under the substitution $x\rightarrow x-1$ we have:
$$f(\frac{1}{x-1})-f(x)=x-1$$ $$-f(x)=x-1-f(\frac{1}{x-1})$$ $$f(x)=-x+1+f(\frac{1}{x-1})$$ $$f(x)=-x+O(1)$$
Where $\lim_{x\to\infty}1+f(\frac{1}{x-1})=1+f(0)=O(1)$ because by assumption $f$ is analytic at $0$ and therefore continuous at $0$, so we are able to interchange the limits.
A few hints that might help...
- $1/x = x+1$ when $x = \frac{\pm\sqrt{5}-1}2$
- Differentiating gives: $-\frac{f'(1/x)}{x^2}+f'(1+x)=1$
- Differentiating again gives: $f''(1+x)+\frac{f''(1/x)}{x^4}+\frac{2f'(1/x)}{x^3}=0$ - this can then be continued.
- An "analytic function" has a Taylor series at any point that is convergent within a non-zero region around the point. So what would the Taylor series look like at the points given in hint 1?
ADDED:
A consideration of limits may also be useful. Indeed, with a substitution of $x=1/y-1$, you have $$f\left(\frac{y}{1-y}\right)+f\left(\frac1y\right)=\frac1y-1$$
We can then cancel out the $\frac1y$ term by first replacing $y$ with $x$, and limits from here may be useful.